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3 years ago

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A circuit consists of two very low-resistance rails separated by 0.300 m, with a 50.0 ohm resistor connected across them at one

end and a conducting, moveable bar at the other end. The circuit is in a uniform 0.100 T magnetic field that is parallel to the area vector of the circuit. The bar is moving at a constant velocity such that the resistor dissipates 1.50 W of power. What is the speed of the bar

1 answer:

Levart [38]3 years ago

6 0

Answer:

v = 288.67 m/s

Explanation:

Given that:

length of the separation I = 0.3

Resistance R = 50.0 ohms

Magnetic field B = 0.100 T

Power dissipated = 1.50 W

In a simple circuit, The Emf voltage of the power dissipated can be determined via the expression:

Power = $\frac{V^2}{R}$

1.5 = $\frac{V^2}{50}$

$V^2 = 1.5*50$

$V^2 = 75$

$V = \sqrt{75}$

V = 8.660 V

From the Electromotive Force Emf ; we can calculate the speed of the bar

i.e Emf = B×v×l

8.660 = 0.1 × v × 0.3

v = $\frac {8.66}{0.1*0.3}$

v = 288.67 m/s

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The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc

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Answer:

The image distance is 20.0 cm.

Explanation:

Given that,

Power = 1.55 dp

Distance between book to eye = 26.0+3.00=29.0 cm

We need to calculate the focal length

Using formula of focal length

$f = \dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{1}{1.55}$

$f=0.645\ m$

$f=64.5\ cm$

We need to calculate the image distance

Using lens formula

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{-u}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{64.5}-\dfrac{1}{-29}$

$\dfrac{1}{v}=\dfrac{187}{3741}$

$v=20.0\ cm$

Hence, The image distance is 20.0 cm.

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3 years ago

A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.

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A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

a)

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

$W=mg$ : weight of the ladder, with m = 20 kg (mass) and $g=9.8 m/s^2$ (acceleration of gravity)

$W_M=Mg$ : weight of the person, with M = 54 kg (mass)

$N_1$ : normal reaction exerted by the wall on the ladder

$N_2$ : normal reaction exerted by the floor on the ladder

$F_f = \mu N_2$ : force of friction between the floor and the ladder, with $\mu$ (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

$W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}$

where

$Wsin 21^{\circ}$ is the component of the weight of the ladder perpendicular to the ladder

$W_M sin 21^{\circ}$ is the component of the weight of the man perpendicular to the ladder

$N_1 sin 69^{\circ}$ is the component of the normal force perpendicular to the ladder

And solving for $N_1$ , we find the force exerted by the wall on the ladder:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N$

B)

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of $N_2$ .

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

$\sum F_y = 0\\N_2 - W - W_M =0$

And substituting and solving for N2, we find:

$N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N$

C)

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

$\sum F_x = 0\\F_f - N_1 = 0$

And re-writing the equation,

$\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257$

So, the minimum value of the coefficient of friction is 0.257.

D)

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

$\mu = \frac{N_1}{N_2}$

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}$

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)$

And substituting, we get

$N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N$

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

$\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332$

Learn more about torques and equilibrium:

brainly.com/question/5352966

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3 years ago

Coal and other fossil fuels are considered to be "nonrenewable" sources of energy. How does the use of nonrenewable fuels affect

nignag [31]

Answer: D Although the total energy remains constant, nonrenewable fuels convert chemical energy into forms that are difficult or impossible to use again.

Explanation:

The first law of thermodynamics says that energy can neither be created nor destroyed; energy can only be transferred or changed from one form to another.

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Which of the following frictionless ramps (A, B, or C) will give the ball the greatest speed at the bottom of the ramp? Explain.

masya89 [10]

The velocity would be the same for all ramps.

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3 years ago

Read 2 more answers

The captain of a boat wants to travel directly across a river that flows due exst with a speed of 100 m/sHe starts from the sout

Salsk061 [2.6K]

Answer:

86.51° North of West or 273.49°

Explanation:

Let V' = velocity of boat relative to the earth, v = velocity of boat relative to water and V = velocity of water.

Now, by vector addition V' = V + v'.

Since v' = 6.10 m/s in the north direction, v' = (6.10 m/s)j and V = 100 m/s in the east direction, V = (100 m/s)i. So that

V' = V + v'

V' = (100 m/s)i + (6.10 m/s)j

So, we find the direction,Ф the boat must steer to from the components of V'.

So tanФ = 6.10 m/s ÷ 100 m/s

tanФ = 0.061

Ф = tan⁻¹(0.061) = 3.49°

So, the angle from the north is thus 90° - 3.49° = 86.51° North of West or 270° + 3.49° = 273.49°

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3 years ago